91Ó°ÊÓ

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther (Reference: Hummingbirds, K. Long and W. Alther). A small group of 15 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is \(\bar{x}=3.15\) grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with \(\sigma=0.33\) gram. (a) Find an \(80 \%\) confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (b) What conditions are necessary for your calculations? (c) Give a brief interpretation of your results in the context of this problem. (d) Sample Size Find the sample size necessary for an \(80 \%\) confidence level with a maximal error of estimate \(E=0.08\) for the mean weights of the hummingbirds.

Step by step solution

01

Understanding the Problem

We are given that a group of 15 Allen's hummingbirds has an average weight \(\bar{x} = 3.15\) grams, with a known population standard deviation \(\sigma = 0.33\) grams. The task is to calculate an 80% confidence interval for the average weight and determine the sample size for a specific error margin in the estimate.

02

Determine the Z-score for 80% Confidence Level

For an 80% confidence interval, the critical z-value can be found using the standard normal distribution table. The 80% confidence level implies 10% in each tail (1 - 0.8 = 0.2), so we look for the z-value at 90% (0.9) cumulative probability, which is approximately \(z = 1.28\).

03

Calculate the Confidence Interval

The formula for the confidence interval is \(\bar{x} \pm z \frac{\sigma}{\sqrt{n}}\), where \(n = 15\). Substituting the values, we get:\[3.15 \pm 1.28 \frac{0.33}{\sqrt{15}}\]. Calculate \(\frac{0.33}{\sqrt{15}} \approx 0.0852\), and multiply by 1.28 to find the margin of error, \(E = 0.1089\). Therefore, the confidence interval is \((3.15 - 0.1089, 3.15 + 0.1089)\) or \( (3.0411, 3.2589) \).

04

List Conditions for Calculations

The primary requirements are: (1) the sample is randomly selected, (2) the sample size is less than 5% of the population (ensuring independence), and (3) the population follows a normal distribution or the sample size is large enough by the Central Limit Theorem.

05

Interpret the Confidence Interval

This confidence interval means that we are 80% confident that the true average weight of Allen's hummingbirds in the study region lies between 3.0411 and 3.2589 grams.

06

Calculate Sample Size for a Given Error Margin

To find the sample size \(n\) necessary for a maximal error of estimate \(E = 0.08\) and an 80% confidence level with \(z = 1.28\), use the formula \(n = \left(\frac{z \sigma}{E}\right)^2\). Substituting in the values gives:\[n = \left(\frac{1.28 \times 0.33}{0.08}\right)^2 = \left(\frac{0.4224}{0.08}\right)^2 = 27.9025\]. Rounding up gives \(n = 28\).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution

Normal distribution is a fundamental concept in statistics that describes how data is spread around the mean. It's often called a "bell curve" due to its characteristic shape, which is symmetric and peaks at the mean. Here are a few important points to understand about normal distribution:

• Most values cluster around the mean, and as we move further away, the number of occurrences decreases.
• It's defined by its mean (average) and standard deviation (a measure of spread).
• In a normal distribution, about 68% of data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.

In the context of the hummingbird weight study, we assume that the weights are normally distributed for accurate calculation of the confidence interval. This assumption allows us to use statistical methods, such as finding the Z-score, confidently.

Sample Size

Sample size is the number of observations or data points you include in your study. The size of your sample has a direct impact on the reliability and accuracy of your statistical estimates. Key things to understand about sample size include:

• Larger samples tend to give more accurate estimations of the population parameter.
• The sample should be random to avoid bias and ensure representativeness of the population.
• The sample size is related to the margin of error. A smaller sample size increases the margin of error, while a larger sample decreases it.

In our scenario, starting with a sample size of 15, calculations showed that for better precision (i.e., smaller margin of error of 0.08), a sample size of 28 would be necessary. This increase gives us more confidence in accurately reflecting the true average weight of the hummingbirds.

Z-score

The Z-score quantifies the number of standard deviations a particular value is from the mean. It is an essential component in calculating confidence intervals. Here's what you need to know about Z-scores:

• A Z-score of 0 indicates that the data point's score is identical to the mean.
• Z-scores can be positive (above the mean) or negative (below the mean).
• They help us determine how unusual a particular measurement is and allow us to compute probabilities.

For the hummingbird study, a Z-score of 1.28 is used to find the 80% confidence interval. This score corresponds to the level of confidence, indicating that 80% of intervals formed this way will contain the true mean weight.

Margin of Error

The margin of error is an indicator of the amount by which your sample's estimate may differ from the true population parameter. It reflects the precision of your sample estimates. Important aspects of margin of error include:

• A smaller margin of error indicates a more precise estimate.
• It is influenced by both the standard deviation and sample size; larger samples and lower variance lead to smaller margins.
• The margin of error is expressed in the same unit as the data; it complements the confidence interval.

In the stated problem, a calculated margin of error of approximately 0.1089 grams implies that the true mean weight is expected to not deviate more than this from the sample mean in 80% of cases. Adjusting sample size and confidence level can help manage and reduce this margin for more precise insights.